Công thức nhân đôi, nhân ba, hạ bậc - Toán 11

\(\sin 2a = 2\sin a\cos a\);

\(\cos 2a = {\cos ^2}a - {\sin ^2}a = 2{\cos ^2}a - 1 = 1 - 2{\sin ^2}a\);

\(\tan 2a = \frac{{2\tan a}}{{1 - {{\tan }^2}a}}\).

\(\sin 3a = 3\sin a - 4{\sin ^3}a\);

\(\cos 3a = 4{\cos ^3}a - 3\cos a\);

\(\tan 3a = \frac{{3\tan a - {{\tan }^3}a}}{{1 - 3{{\tan }^2}a}}\).

\({\cos ^2}a = \frac{{1 + \cos 2a}}{2}\);

\({\sin ^2}a = \frac{{1 - \cos 2a}}{2}\);

\({\sin ^3}a = \frac{{3\sin a - \sin 3a}}{4}\);

\({\cos ^3}a = \frac{{3\cos a + \cos 3a}}{4}\).

1) Cho \(\cos a =  - \frac{1}{3}\) \(\left( {\frac{\pi }{2} < a < \pi } \right)\). Tính sin2a.

Giải:

Vì \(\frac{\pi }{2} < a < \pi \) nên sina > 0.

Do đó \(\sin a = \sqrt {1 - {{\cos }^2}a}  = \sqrt {1 - {{\left( { - \frac{1}{3}} \right)}^2}}  = \sqrt {\frac{8}{9}}  = \frac{{2\sqrt 2 }}{3}\).

Vậy \(\sin 2a = 2\sin a\cos a = 2.\frac{{2\sqrt 2 }}{3}.\left( { - \frac{1}{3}} \right) =  - \frac{{4\sqrt 2 }}{9}\).

2) Cho \(\sin a + \cos a = \frac{1}{2}\). Tính:

a) sin2a;

b) cos4a.

Giải:

a) \(\sin a + \cos a = \frac{1}{2}\)

\( \Rightarrow {\left( {\sin a + \cos a} \right)^2} = \frac{1}{4}\)

\( \Leftrightarrow {\sin ^2}a + {\cos ^2}a + 2\sin a\cos a = \frac{1}{4}\)

\( \Leftrightarrow 1 + 2\sin a\cos a = \frac{1}{4}\)

\( \Leftrightarrow 2\sin a\cos a = \frac{1}{4} - 1 =  - \frac{3}{4}\).

b) \(\cos 4a = \cos (2.2a) = 1 - {\sin ^2}2a =  - \frac{1}{8}\).

3) Biết \(\cos \frac{\pi }{6} = \frac{{\sqrt 3 }}{2}\). Tính \(\cos \frac{\pi }{{12}}\).

Giải:

Ta có \({\cos ^2}\frac{\pi }{{12}} = \frac{{1 + \cos \frac{\pi }{6}}}{2} = \frac{{2 + \sqrt 3 }}{4}\).

Mà \(\cos \frac{\pi }{{12}} > 0\) nên \(\cos \frac{\pi }{{12}} = \frac{{\sqrt {2 + \sqrt 3 } }}{2}\).